IB Math AA HL Binomial Theorem & Series — Complete Cheatsheet

The Binomial Theorem is one of the most reliable mark-earners in IB Mathematics Analysis & Approaches HL — but only if you know where the traps are. The SL version (positive integer $n$, finite expansion) is heavily tested in Paper 1 and almost always asks for a specific term using the general term $T_{r+1}$. The AHL extension (rational or negative $n$) introduces the generalised binomial series, which links directly to the Maclaurin series, partial fractions, and the geometric series — and brings the brutal validity condition $|x| < 1$ that costs students a guaranteed mark when forgotten.

This cheatsheet condenses every formula, identity, technique, trick and trap from SL 1.3 and AHL 1.9 into one revision page. Scroll to the bottom for the printable PDF, the full Notes, the worked tutorials and the marked-up solutions in the gated student library.

§1 — Combinations & the Binomial Theorem SL 1.3

The binomial coefficient

$$\binom{n}{r} = C^n_r = \dfrac{n!}{r!\,(n-r)!}$$

• $\binom{n}{0} = \binom{n}{n} = 1$
• $\binom{n}{r} = \binom{n}{n-r}$   (symmetry)
• Pascal: $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$

P1 step-back method (no calculator)

Rule: Start at $n$, multiply $r$ factors downward, divide by $r!$.

Examples: $\binom{10}{2} = \dfrac{10 \times 9}{2} = 45$   ·   $\binom{8}{3} = \dfrac{8 \times 7 \times 6}{6} = 56$.

Pascal's triangle

Row $n$ gives the coefficients of $(a + b)^n$. Sum of row $n$ $= 2^n$.

Key identities

Sum of even-indexed $=$ sum of odd-indexed $= 2^{n-1}$.

§2 — The Binomial Theorem & General Term SL 1.3

$$(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r, \qquad n \in \mathbb{N}$$

General term (most important formula):

$$\boxed{\;T_{r+1} = \binom{n}{r}\, a^{n-r}\, b^r\;}$$

Properties of the expansion:

• Exactly $n + 1$ terms.
• Powers of $a$ decrease $n \to 0$.
• Powers of $b$ increase $0 \to n$.
• Every term: $(n - r) + r = n$. ✓
• Coefficients are symmetric.

5-step attack: finding a specific term

1. Write $T_{r+1} = \binom{n}{r}\, a^{n-r}\, b^r$ with your $a$, $b$, $n$ substituted.
2. Collect the power of $x$ — write it as a function of $r$.
3. Set that power equal to what you need (e.g. 3 for $x^3$, 0 for the constant term).
4. Solve for $r$ — must be a non-negative integer $\leq n$.
5. Substitute back and compute (including $\binom{n}{r}$, numerical factors, sign).

Worked micro-examples

Find the term in $x^3$ in $(2x - 1)^7$: $a = 2x$, $b = -1$, $n = 7$.
$T_{r+1} = \binom{7}{r}(2x)^{7-r}(-1)^r = \binom{7}{r}\, 2^{7-r}(-1)^r\, x^{7-r}$.
Set $7 - r = 3 \Rightarrow r = 4$: $T_5 = \binom{7}{4} \cdot 2^3 \cdot (-1)^4 \cdot x^3 = 35 \cdot 8 \cdot 1 \cdot x^3 = 280 x^3$.

Constant term in $\left(x^2 - \dfrac{2}{x}\right)^6$:
$T_{r+1} = \binom{6}{r}(x^2)^{6-r}(-2/x)^r = \binom{6}{r}(-2)^r x^{12 - 3r}$.
Set $12 - 3r = 0 \Rightarrow r = 4$: $T_5 = \binom{6}{4}(-2)^4 = 15 \times 16 = 240$.

Top 6 traps — SL Binomial

§3 — Generalised Binomial Series AHL 1.9

For $n \in \mathbb{Q}$ (any rational), valid for $|x| < 1$:

$$(1 + x)^n = 1 + nx + \dfrac{n(n-1)}{2!} x^2 + \dfrac{n(n-1)(n-2)}{3!} x^3 + \cdots$$

With $ax$ substitution:

$$(1 + ax)^n = 1 + n(ax) + \dfrac{n(n-1)}{2!}(ax)^2 + \cdots, \quad \text{valid for } |x| < \dfrac{1}{|a|}$$

Key differences from SL:

• Series is infinite (never terminates unless $n \in \mathbb{N}_0$).
• Only valid for $|x| < 1$ (or $|ax| < 1$).
• Generalised $\binom{n}{r} = \dfrac{n(n-1) \cdots (n-r+1)}{r!}$ — can be fractional or negative.

Generalised coefficients table

Standard series to memorise ($|x| < 1$)

§4 — The Factoring Trick: $(a + bx)^n$ when $a \neq 1$ AHL 1.9

1. Step 1: Factor out $a^n$:   $(a + bx)^n = a^n\!\left(1 + \dfrac{b}{a} x\right)^n$.
2. Step 2: Expand $(1 + u)^n$ with $u = \dfrac{b}{a} x$.
3. Step 3: State validity $|u| < 1$, i.e. $|x| < \dfrac{|a|}{|b|}$.

Example: Expand $\sqrt{4 + x}$ to 3 terms.
$\sqrt{4 + x} = 2\!\left(1 + \tfrac{x}{4}\right)^{1/2} = 2\!\left[1 + \tfrac{1}{2}\cdot \tfrac{x}{4} + \tfrac{(1/2)(-1/2)}{2}\cdot \tfrac{x^2}{16} + \cdots\right] = 2 + \tfrac{x}{4} - \tfrac{x^2}{64} - \cdots$, valid for $|x| < 4$.

Approximation by substitution

Goal: Use your series to approximate $\sqrt[k]{m}$ or $(1.03)^{-5}$, etc.

1. Expand $(1 + ax)^n$ as a series.
2. Set $1 + ax =$ target base, solve for $x$.
3. Check $|x| < \dfrac{1}{|a|}$ — must be in the valid range.
4. Substitute $x$ into the series and compute.

Top 6 traps — AHL Binomial

§5 — Partial Fractions + Binomial Series AHL 1.9

Method

1. Split into partial fractions: $\dfrac{A}{1 + ax} + \dfrac{B}{1 + bx}$.
2. Expand each using $(1 + u)^{-1} = 1 - u + u^2 - \cdots$.
3. Add the two series; collect powers of $x$.
4. Validity = the more restrictive of the two conditions.

Validity reminder

Worked example

Expand $\dfrac{5}{(1 + x)(1 - 2x)}$ to $x^2$.

Partial fractions: $\dfrac{A}{1 + x} + \dfrac{B}{1 - 2x}$.   $5 \equiv A(1 - 2x) + B(1 + x)$.
$x = -1$: $5 = 3A \Rightarrow A = \tfrac{5}{3}$.   $x = \tfrac{1}{2}$: $5 = \tfrac{3B}{2} \Rightarrow B = \tfrac{10}{3}$.

$\tfrac{5}{3}(1 - x + x^2 - \cdots) + \tfrac{10}{3}(1 + 2x + 4x^2 + \cdots)$
$= 5 + \tfrac{-5 + 20}{3} x + \tfrac{5 + 40}{3} x^2 + \cdots = 5 + 5x + 15x^2 + \cdots$.

Valid for $|x| < \tfrac{1}{2}$.

Validity table

§6 — Connection to Maclaurin & $\ln$ AHL 1.9, AHL 5.19

The generalised binomial series is the Maclaurin series of $(1 + x)^n$:

$f(x) = (1 + x)^n \Rightarrow f^{(r)}(0) = n(n-1) \cdots (n-r+1)$
$f(x) = \displaystyle\sum_{r=0}^{\infty} \dfrac{f^{(r)}(0)}{r!} x^r = \sum_{r=0}^{\infty} \binom{n}{r} x^r$.

From $(1 + x)^{-1}$, integrate to get $\ln$

$\dfrac{d}{dx} \ln(1 + x) = \dfrac{1}{1 + x} = (1 + x)^{-1} = 1 - x + x^2 - x^3 + \cdots$
$\Rightarrow \ln(1 + x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} + \cdots$,   $|x| \leq 1$, $x \neq -1$.

§7 — GDC Skills (Paper 2) Paper 2

§8 — Exam Attack Plan SL 1.3, AHL 1.9

Worked Example — IB-Style Generalised Series

Common Student Questions